TITLE: Degree of a morphism of projective curves
QUESTION [6 upvotes]: I think this is going to be a silly question. I'm happy with the following fact:
If $\alpha : X \to Y$ is a non-constant morphism of irreducible curves, then $\alpha$ induces an embedding of field $k(Y) \hookrightarrow k(X)$ such that $[ k(X) : k(Y) ] = \mathrm{deg}(\alpha)$ is finite.
I have the following question:
Let $\phi = (1:f) : \mathbb P^1 \to \mathbb P^1 $ be a morphism given by a non-constant polynomial $f \in k[t] \subset k(\mathbb P^1)$. Prove that $\mathrm{deg}(\phi) = \mathrm{deg}\ f$.
I can't see why this is true. $\phi$ is a non-constant morphism of smooth irreducible curves, right? Why isn't $\mathrm{deg}(\phi) = 1$? (both domain and codomain have the same function field...)
Thanks

REPLY [3 votes]: It is absolutely true that the domain and codomain have abstractly isomorphic function fields, but the map $\phi$ provides a way of embedding the one function field as a subfield of the other. Let us take a simple example, where $f(t) = t^2$. In this case, the function fields of the domain and codomain are both abstractly isomorphic to $k(x)$, but the embedding of function fields induced by $f$ is $k(t^2)\subset k(t)$. For general polynomial $f$, the embedding of functions fields will be $k(f(t))\subset k(t)$. You should be able to derive what you want from this.